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The solution to be tested in this experiment must be buffered to ph of 10. If the ph were considerably higher, such as 12, how would the [oh-] change

Sagot :

For a solution to be tested in this experiment and must be buffered to pH of 10,  the [OH-] change will be, The hydroxide concentration would be higher and the formation of insoluble hydroxide salts with Mg^2 and Ca^2 would cause the determined concentration of water hardness to be too low.

What would be the [OH-] change?

Generally, the equation for the  Total hardness is mathematically given as

[tex]T=\frac{A*D*1000}{B}[/tex]

Where

[tex]D=\frac{molarity}{0.01}[/tex]

D=0.02/0.01=2

Therefore

[tex]T=\frac{A*D*1000}{B}\\\\T=\frac{20.4*2*1000}{25}[/tex]

T=1632ppm

in conclusion, The hydroxide concentration would be higher .

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The solution to be tested in this experiment must be buffered to a pH of 10. If the pH is high to 12, the change of [OH-] is 1632 ppm.

What is OH- ion?

OH- ion is the hydroxide ion or hydroxyl ion. It is made up of a hydrogen and oxygen ion with a covalent bond.

[tex]A =\dfrac{A \times D \times 1000}{T} \\[/tex]

The buffered pH is 10

The higher pH is 12

D = molarity divided by 0.01, that is the pH

[tex]D = \dfrac{molarity }{0.01}\\\\D = \dfrac{0.02 }{0.01} = 2[/tex]

Now, putting the values in the first equation

[tex]A =\dfrac{20.4 \times 2 \times 1000}{25} = 1632 \;ppm[/tex]

Thus, the correct option is the change in hydroxide ion is 1632 ppm.

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